Bpt class 10

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• Number Systems • Real Numbers • • • • • • • • • Algebra • Pair of Linear Equations in Two Variables • • • • • • • • • • • • • Arithmetic Progressions • • • • • • • • Quadratic Equations • • • • • • • • Polynomials • • • • • • Geometry • Circles • • • • • Triangles • • • • • • • • • • • • • Constructions • • • • Trigonometry • Heights and Distances • • Trigonometric Identities • • • Introduction to Trigonometry • • • • • • • • • • Statistics and Probability • Probability • • • • • • • Statistics • • • • • • • • Coordinate Geometry • Lines (In Two-dimensions) • • • • • • • • Mensuration • Areas Related to Circles • • • • • • Surface Areas and Volumes • • • • • • • • • Internal Assessment Theorem: If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion. Given: In D ABC line l || line BC and line l intersects AB and AC in point P and Q respectively To prove: `"AP"/"PB"="AQ"/"QC"` Construction: Draw seg PC and seg BQ Proof: Δ APQ and Δ PQB have equal heights. `therefore (A(triangle APQ))/(A(triangle PQB))="AP"/"PB"` .....................(I) (areas proportionate to bases) `therefore (A(triangle APQ))/(A(triangle PQC))="AQ"/"QC"`................... (II) (areas proportionate to bases) seg PQ is a common to base of Δ PQB and ΔPQC. seg PQ || seg BC, hence Δ PQB and Δ PQC have equal heights. `A(triangle PQB)=A(triangle PQC)` ................(III) `(A(triangle APQ))/(A(triangle PQB))=(A(...

Transcript Theorem 6.3 (AAA Criteria) If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangle are similar.Given: Two triangles ∆ABC and ∆DEF such that ∠A = ∠D, ∠B = ∠E & ∠C = ∠F To Prove: ∆ABC ~ ∆DEF Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof: In ∆ABC and ∆DPQ AB = DP ∠A = ∠D AC = DQ ⇒ ∆ABC ≅ ∆DPQ ⇒ ∠B = ∠P But, ∠B = ∠E Thus, ∠P = ∠E For lines PQ & EF with transversal PE, ∠ P & ∠ E are corresponding angles, and they are equal Hence, PQ ∥ EF. Since, PQ ∥ EF. By Theorem 6.1, : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹 𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄 Adding 1 on both sides. 𝑃𝐸/𝐷𝑃 + 1 = 𝑄𝐹/𝐷𝑄 + 1 (𝑃𝐸 + 𝐷𝑃)/𝐷𝑃 = (𝑄𝐹 + 𝐷𝑄)/𝐷𝑄 𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄 ⇒ 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹 And by construction DP = AB and DQ = AC ⇒ 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 Similarly, we can prove that 𝐴𝐵/𝐷𝐸 = 𝐵𝐶/𝐸𝐹 Therefore, 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 = 𝐵𝐶/𝐸𝐹 Since all 3 sides are in proportion ∴ ∆ABC ~ ∆DEF Hence Proved. Show More

Theorem 6.1 Thale’s Theorem/ Basic Proportionality Theorem (BPT) Statement: ” If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points. The other two sides are divided in the same ratio.” OR Prove that in a triangle a line parallel to one side divides remaining two sides in the same ratio Given: In Δ ABC DE // ( parallel to ) BC To Prove: AD/BD = AE/CE Construction: DF ⊥ ( Perpendicular ) on AC EG ⊥ ( Perpendicular ) on AB Join B to E and C to D Proof: According to figure: In Δ ADE and Δ BDE have common altitude (Perpendicular) = EG ∴ area Δ ADE /area Δ BDE = 1/2×b×h / 1/2×b×h ⇒ area Δ ADE /area Δ BDE = AD×EG / BD×EG ⇒ area Δ ADE /area Δ BDE = AD / BD————-equation (i) Similarly In Δ ADE and Δ CED Have common altitude(perpendicular) = DF ar Δ ADE / ar Δ CED = 1/2×b×h / 1/2×b×h ⇒ ar Δ ADE / ar Δ CED = AE×DF / CE×DF ⇒ ar Δ ADE / ar Δ CED = AE / CE ————-equation (ii) Δ BDE and Δ CED are lies on Common base (DE) and between same parallel lines ∴ ar Δ BDE = ar Δ CED Dividing both side by area Δ ADE ⇒ ar Δ BDE/ar Δ ADE = ar Δ CED/ar Δ ADE ⇒ ar Δ ADE/ar Δ BDE = ar Δ ADE/ar Δ CED [ After reciprocal of both side] ————–equation(iii) By comparing equation (i) and equation(ii) with equation (iii) We get AD/BD = AE/CE Hence proved Post navigation

10th Triangle (Similarity) : Problems and solution toexcelin exam Similar figures: “Two similar figures have the same shape but not necessarily the same sizes are called similar figures. “ This verifies that congruent figures are similar but the similar figures need not be congruent. Conditions for similarity of polygon: Two polygons of the same number of sides are similar, if (i) Their corresponding angles are equal and (ii) Their corresponding sides are in the same ratio (or proportion). Note: The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons. Equiangular triangles: If corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: “The ratio of any two corresponding sides in two equiangular triangles is always the same.” Q. The Basic Proportionality Theorem (now known as the Thales Theorem) : “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. “ [Prove it.] Q. The converse of The Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [Prove it by contradiction methods] Q. In a triangle ABC, E and F are point on AB and AC and EF || BC. Prove that AB/AE = AC/A...

Theorem 6.1 Thale’s Theorem/ Basic Proportionality Theorem (BPT) Statement: ” If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points. The other two sides are divided in the same ratio.” OR Prove that in a triangle a line parallel to one side divides remaining two sides in the same ratio Given: In Δ ABC DE // ( parallel to ) BC To Prove: AD/BD = AE/CE Construction: DF ⊥ ( Perpendicular ) on AC EG ⊥ ( Perpendicular ) on AB Join B to E and C to D Proof: According to figure: In Δ ADE and Δ BDE have common altitude (Perpendicular) = EG ∴ area Δ ADE /area Δ BDE = 1/2×b×h / 1/2×b×h ⇒ area Δ ADE /area Δ BDE = AD×EG / BD×EG ⇒ area Δ ADE /area Δ BDE = AD / BD————-equation (i) Similarly In Δ ADE and Δ CED Have common altitude(perpendicular) = DF ar Δ ADE / ar Δ CED = 1/2×b×h / 1/2×b×h ⇒ ar Δ ADE / ar Δ CED = AE×DF / CE×DF ⇒ ar Δ ADE / ar Δ CED = AE / CE ————-equation (ii) Δ BDE and Δ CED are lies on Common base (DE) and between same parallel lines ∴ ar Δ BDE = ar Δ CED Dividing both side by area Δ ADE ⇒ ar Δ BDE/ar Δ ADE = ar Δ CED/ar Δ ADE ⇒ ar Δ ADE/ar Δ BDE = ar Δ ADE/ar Δ CED [ After reciprocal of both side] ————–equation(iii) By comparing equation (i) and equation(ii) with equation (iii) We get AD/BD = AE/CE Hence proved Post navigation

• Number Systems • Real Numbers • • • • • • • • • Algebra • Pair of Linear Equations in Two Variables • • • • • • • • • • • • • Arithmetic Progressions • • • • • • • • Quadratic Equations • • • • • • • • Polynomials • • • • • • Geometry • Circles • • • • • Triangles • • • • • • • • • • • • • Constructions • • • • Trigonometry • Heights and Distances • • Trigonometric Identities • • • Introduction to Trigonometry • • • • • • • • • • Statistics and Probability • Probability • • • • • • • Statistics • • • • • • • • Coordinate Geometry • Lines (In Two-dimensions) • • • • • • • • Mensuration • Areas Related to Circles • • • • • • Surface Areas and Volumes • • • • • • • • • Internal Assessment Theorem: If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion. Given: In D ABC line l || line BC and line l intersects AB and AC in point P and Q respectively To prove: `"AP"/"PB"="AQ"/"QC"` Construction: Draw seg PC and seg BQ Proof: Δ APQ and Δ PQB have equal heights. `therefore (A(triangle APQ))/(A(triangle PQB))="AP"/"PB"` .....................(I) (areas proportionate to bases) `therefore (A(triangle APQ))/(A(triangle PQC))="AQ"/"QC"`................... (II) (areas proportionate to bases) seg PQ is a common to base of Δ PQB and ΔPQC. seg PQ || seg BC, hence Δ PQB and Δ PQC have equal heights. `A(triangle PQB)=A(triangle PQC)` ................(III) `(A(triangle APQ))/(A(triangle PQB))=(A(...

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Transcript Theorem 6.3 (AAA Criteria) If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangle are similar.Given: Two triangles ∆ABC and ∆DEF such that ∠A = ∠D, ∠B = ∠E & ∠C = ∠F To Prove: ∆ABC ~ ∆DEF Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ. Proof: In ∆ABC and ∆DPQ AB = DP ∠A = ∠D AC = DQ ⇒ ∆ABC ≅ ∆DPQ ⇒ ∠B = ∠P But, ∠B = ∠E Thus, ∠P = ∠E For lines PQ & EF with transversal PE, ∠ P & ∠ E are corresponding angles, and they are equal Hence, PQ ∥ EF. Since, PQ ∥ EF. By Theorem 6.1, : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 𝐷𝑃/𝑃𝐸 = 𝐷𝑄/𝑄𝐹 𝑃𝐸/𝐷𝑃 = 𝑄𝐹/𝐷𝑄 Adding 1 on both sides. 𝑃𝐸/𝐷𝑃 + 1 = 𝑄𝐹/𝐷𝑄 + 1 (𝑃𝐸 + 𝐷𝑃)/𝐷𝑃 = (𝑄𝐹 + 𝐷𝑄)/𝐷𝑄 𝐷𝐸/𝐷𝑃 = 𝐷𝐹/𝐷𝑄 ⇒ 𝐷𝑃/𝐷𝐸 = 𝐷𝑄/𝐷𝐹 And by construction DP = AB and DQ = AC ⇒ 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 Similarly, we can prove that 𝐴𝐵/𝐷𝐸 = 𝐵𝐶/𝐸𝐹 Therefore, 𝐴𝐵/𝐷𝐸 = 𝐴𝐶/𝐷𝐹 = 𝐵𝐶/𝐸𝐹 Since all 3 sides are in proportion ∴ ∆ABC ~ ∆DEF Hence Proved. Show More

10th Triangle (Similarity) : Problems and solution toexcelin exam Similar figures: “Two similar figures have the same shape but not necessarily the same sizes are called similar figures. “ This verifies that congruent figures are similar but the similar figures need not be congruent. Conditions for similarity of polygon: Two polygons of the same number of sides are similar, if (i) Their corresponding angles are equal and (ii) Their corresponding sides are in the same ratio (or proportion). Note: The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons. Equiangular triangles: If corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: “The ratio of any two corresponding sides in two equiangular triangles is always the same.” Q. The Basic Proportionality Theorem (now known as the Thales Theorem) : “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. “ [Prove it.] Q. The converse of The Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [Prove it by contradiction methods] Q. In a triangle ABC, E and F are point on AB and AC and EF || BC. Prove that AB/AE = AC/A...